Constructions
Division of a line segment in a given ratiovertical-align: middle;
Example:
Draw and divide it in the ratio 2:5. Justify your construction.
Solution:
Steps of construction
Draw = 9 cm
Draw a ray , making an acute angle with PQ.
Mark 7 (= 2 + 5) points A1, A2, A3 … A7 along PX such that
PA1 = A1A2 = A2 A3 = A3 A4 = A4 A5 = A5 A6 = A6 A7
Join QA7
Through the point A2, draw a line parallel to A7Q by making an angle
equal to ∠PA7Q at A2, intersecting PQ at point R.
PR:RQ = 2:5
Justification:
We have A2R || A7Q
Construction of a triangle similar to a given triangle as per the given scale factor
Case I: Scale factor less than 1
Example:
Draw a ΔABC with sides BC = 8 cm, AC = 7 cm, and ÐB = 70°. Then,
construct a similar triangle whose sides are 35th of the corresponding sides
of ΔABC.
Justify your construction.
Solution:
Steps of construction:
Draw BC = 8 cm
At B, draw ∠XBC = 70°
With C as centre and radius 7 cm, draw an arc intersecting BX at A.
Join AB, and DABC is thus obtained.
Draw a ray BY→ , making an acute angle with BC.
Mark 5 points, B1, B2, B3, B4, B5, along BY such that
BB1 = B1B2 = B2B3 = B3B4 = B4B5
Join CB5
Through the point B3, draw a line parallel to B5 C by making an angle
equal to ∠BB5C, intersecting BC at C´.
Through the point C´, draw a line parallel to AC, intersecting BA at A´.
Thus, ΔA´BC´ is the required triangle.
Justification:
By construction, we have
Now, in ΔABC,
Since, AC || A'C' , so
Case II: Scale factor greater than 1
Example:
Construct an isosceles triangle with base 5 cm and equal sides of 6 cm.
Then, construct another triangle whose sides are 43rdof the corresponding
sides of the first triangle.
Solution:
Steps of construction:
Draw BC = 5 cm
With B and C as the centre and radius 6 cm, draw arcs on the same side of
BC, intersecting at A.
Join AB and AC to get the required ΔABC.
Draw a ray , making an acute angle with BC on the side opposite to the
vertex A.
Mark 4 points B1, B2, B3, B4, along BX such that
BB1 = B1B2 = B2B3 = B3 B4
Join B3 C. Draw a line through B4 parallel to B3 C, making an angle equal
to ∠BB3 C, intersecting the extended line segment BC at C´.
Through point C´, draw a line parallel to CA, intersecting extended BA
at A´.
The resulting ΔA´BC´ is the required triangle.
Construction of tangents to a circle
Example:
Draw a circle of radius 3 cm. From a point 5 cm away from its centre,
construct a pair of tangents to the circle and measure their lengths.
Solution:
Steps of construction:
1.
1.
1. Draw a circle with centre O and radius 3 cm. Take a point P such that OP = 5 cm, and
then join OP.
2. Draw the perpendicular bisector of OP. Let M be the mid point of OP.
3. With M as the centre and OM as the radius, draw a circle. Let it intersect the
previously drawn circle at A and B.
4. Joint PA and PB. Therefore, PA and PB are the required tangents. It can be observed
that PA = PB = 4 cm.
